1 files changed, 54 insertions, 0 deletions
diff --git a/stdlib/ldiv.c b/stdlib/ldiv.c
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index 0000000000..a7796d8e95
--- /dev/null
+++ b/stdlib/ldiv.c
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+/* Copyright (C) 1992, 1997 Free Software Foundation, Inc.
+   This file is part of the GNU C Library.
+
+   The GNU C Library is free software; you can redistribute it and/or
+   modify it under the terms of the GNU Lesser General Public
+   License as published by the Free Software Foundation; either
+   version 2.1 of the License, or (at your option) any later version.
+
+   The GNU C Library is distributed in the hope that it will be useful,
+   but WITHOUT ANY WARRANTY; without even the implied warranty of
+   MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
+   Lesser General Public License for more details.
+
+   You should have received a copy of the GNU Lesser General Public
+   License along with the GNU C Library; if not, write to the Free
+   Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
+   02111-1307 USA.  */
+
+#include <stdlib.h>
+
+
+/* Return the `ldiv_t' representation of NUMER over DENOM.  */
+ldiv_t
+ldiv (long int numer, long int denom)
+{
+  ldiv_t result;
+
+  result.quot = numer / denom;
+  result.rem = numer % denom;
+
+  /* The ANSI standard says that |QUOT| <= |NUMER / DENOM|, where
+     NUMER / DENOM is to be computed in infinite precision.  In
+     other words, we should always truncate the quotient towards
+     zero, never -infinity.  Machine division and remainer may
+     work either way when one or both of NUMER or DENOM is
+     negative.  If only one is negative and QUOT has been
+     truncated towards -infinity, REM will have the same sign as
+     DENOM and the opposite sign of NUMER; if both are negative
+     and QUOT has been truncated towards -infinity, REM will be
+     positive (will have the opposite sign of NUMER).  These are
+     considered `wrong'.  If both are NUM and DENOM are positive,
+     RESULT will always be positive.  This all boils down to: if
+     NUMER >= 0, but REM < 0, we got the wrong answer.  In that
+     case, to get the right answer, add 1 to QUOT and subtract
+     DENOM from REM.  */
+
+  if (numer >= 0 && result.rem < 0)
+    {
+      ++result.quot;
+      result.rem -= denom;
+    }
+
+  return result;
+}